1 / 26

S.4 Mathematics

S.4 Mathematics. y. (3, 4). x + y –7 = 0. x. 0. 2 x – 3 y +6=0. Put (3,4) into x + y –7 =0. Put (3,4) into 2 x –3 y +6 =0. LHS = (2)3 – 3(4) + 6. LHS = 3+4 – 7. = 0. = 0. = RHS. = RHS. (3,4) is the solution of the equations of x + y –7 =0 and

tammy
Download Presentation

S.4 Mathematics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. S.4 Mathematics

  2. y (3, 4) x + y –7 = 0 x 0 2x – 3y +6=0 Put (3,4) into x +y –7 =0 Put (3,4) into 2x –3y +6 =0 LHS = (2)3 – 3(4) + 6 LHS = 3+4 – 7 = 0 = 0 = RHS = RHS (3,4) is thesolutionof the equations of x +y –7 =0 and 2x –3y +6 =0

  3. y x + y –7 = 0 x 0 2x – 3y + 6 = 0 (3, 4) What is the solution of the simultaneous equations?

  4. y x 0 Two points of intersection

  5. y x 0 One point of intersection

  6. y 0 No points of intersection x What is the relationship between the number of pointsof intersection and the value of discriminant?

  7. y y y x x 0 0 0 Case 1:2 points of intersection ∆ > 0 Case 2: 1 point of intersection ∆ = 0 Case 3:No point of intersection ∆ < 0

  8. Example Determine the number of points of intersection of the parabola and the straight line. Parabola: Straight line:

  9. No need to solve the eq. There are two points of intersection

  10. Determine the number of points of intersection of the parabola and the straight line. Parabola: Straight line:

  11. # 1 There is no point of intersection

  12. # 2 There are two points of intersection

  13. # 3 There is one point of intersection

  14. y = – 3 x – 6 No point of intersection

  15. y = 2 x – 4 2 3 – 4 2 Two points of intersection

  16. 2 x – y – 12= 0 2 3 4 – 8 – 6 – 4 One point of intersection

  17. Determine the number of points of intersection of the parabola and the straight line. Parabola: Straight line: We can use : I) Graphical method II) Discriminant method Any other method?

  18. Determine the number of points of intersection of the parabola and the straight line. Parabola: Straight line: We can use : I) Graphical method II) Discriminant method Any other method?

  19. #2 ∴(3,4) and (-2,-8) are the 2 points of intersection.

  20. Which method is the fastest in determining the number of points of intersection of the parabola and the straight line? I) Graphical method II) Discriminant method III) Solving the simultaneous equations (Algebraic method)

  21. Exercise 1. If the parabola y = – x2 + 2x + 5 and the line y = k intersect at one point, find the value of k.

  22. # Ex.1 If the parabola and the line intersect at one point , then the discriminant equals to zero. 24 – 4k = 0 ∴ k = 6

  23. Exercise 2. If the straight line y = 3x + kdoes not cut the parabola y = x2 – 3 x + 2 at any point, find the range of values of k. There is no point of intersection

  24. # Ex.2 There is no point of intersection so the discriminant is less thanzero. 28 + 4k < 0 ∴ k < – 7

  25. Exercise 3. If the straight line 2x – y – 1 = 0 cuts the parabola y = 3 x2 + 5x + k at two points, where k is an integer. Find the largest value of k.

  26. # Ex.3 There are two points of intersection so the discriminant is greater thanzero. ∴ k < – 0.25 – 3 – 12 k > 0 The largest value of k is – 1

More Related